I just discovered that you cannot always use brace initialization when passing arguments to macros. I found this when an ASSERT() macro failed to compile. However, the following example illustrates the problem:

#include <iostream>
#include <string>
using namespace std;

#define PRINT_SIZE( f ) cout << "Size=" << (f).size() << endl;

int main()
{
  PRINT_SIZE( string("ABC") );  // OK, prints: "Size=3"
  PRINT_SIZE( string{"ABC"} );  // OK, prints: "Size=3"

  PRINT_SIZE( string("ABCDEF",3) ); // OK, prints: "Size=3"
  PRINT_SIZE( string{"ABCDEF",3} ); // Error: macro 'PRINT_SIZE' passed 2 arguments, but takes just 1

   return 0;
}

Is there a reason why macros cannot be made to work with brace initialization?

Edit:

I have since discovered that you can also use a variadic macro, and that solves the problem perfectly:

#include <iostream>
#include <string>
using namespace std;

#define PRINT_SIZE( ... ) cout << "Size=" << (__VA_ARGS__).size() << endl;

int main()
{
  PRINT_SIZE( string("ABC") );  // OK, prints: "Size=3"
  PRINT_SIZE( string{"ABC"} );  // OK, prints: "Size=3"

  PRINT_SIZE( string("ABCDEF",3) ); // OK, prints: "Size=3"
  PRINT_SIZE( string{"ABCDEF",3} ); // OK, prints: "Size=3"

  return 0;
}

The list is split into several macro parameters. When you write

PRINT_SIZE( string{"ABCDEF",3} );

This attempts to expand the macro PRINT_SIZE with two parameters, one string{"ABCDEF" and one 3}, which fails. This can be worked around in many cases (including yours) by adding another pair of parentheses:

PRINT_SIZE( (string{"ABCDEF",3}) );

These parentheses prevent the splitting of the argument, so that PRINT_SIZE is expanded with a single argument (string{"ABCDEF",3}) (note that the parentheses are part of the argument).